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Question 1
Correct
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Signals pass through neuromuscular junctions via the neurotransmitter acetylcholine. After release from the skeletal neuromuscular junction, acetylcholine:
Your Answer: Causes postsynaptic depolarisation
Explanation:Acetylcholine is released from the presynaptic membrane into the cleft where it binds to the ion gated channels on the post synaptic membrane, causing them to open. This results in sodium entering into the fibre and further depolarizing it, creating an action potential.
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This question is part of the following fields:
- General
- Physiology
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Question 2
Correct
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A teenage Somalian boy presents with a complaint of an enlarged lower jaw. His blood film shows blast cells and macrophages. Which virus is responsible for this?
Your Answer: Epstein–Barr virus
Explanation:Burkitt’s lymphoma is a type of non-Hodgkin’s lymphoma. Histologically it is characterised by a starry sky appearance due to numerous neoplastic macrophages which are required to clear the rapidly dividing tumour cells/blast cells. Burkitt’s lymphoma commonly affects the jaw bone, forming a huge tumour mass. It is associated with translocation of c-myc gene and has three types: 1) endemic/African type, 2)sporadic and 3)immunodeficiency-associated. The first type is strongly associated with EBV.
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This question is part of the following fields:
- General
- Physiology
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Question 3
Incorrect
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During cardiac catheterisation in a 20-year old man, the following data is obtained: Pressure (mmHg), O2 saturation (%) Right atrium 7 (N = 5) 90 (N = 75), Right ventricle 35/7 (N = 25/5) 90 (N = 75), Pulmonary artery 35/8 (N = 25/15), 90 (N = 75), Left atrium 7 (N = 9) 95 (N = 95), Left ventricle 110/7 (N = 110/9) 95 (N = 95), Aorta 110/75 (N = 110/75) 95 (N = 95) where N = Normal value. What is the likely diagnosis?
Your Answer: Pulmonary stenosis
Correct Answer: Atrial septal defect
Explanation:A congenital heart disease, ASD or atrial septal defect leads to a communication between the right and left atria due to a defect in the interatrial septum. This leads to mixing of arterial and venous blood from the right and left side of the heart. The hemodynamic significance of this defect depends on the presence of shunting of blood. Normally, the left side of the heart has higher pressure than the right as the left side has to pump blood throughout the body. A large ASD (> 9 mm) will result in a clinically significant left-to-right shunt, causing volume overload of the right atrium and ventricle, eventually leading to heart failure. Cardiac catheterization would reveal very high oxygen saturation in the right atrium, right ventricle and pulmonary artery. Eventually, the left-to-right shunt will lead to pulmonary hypertension and increased afterload in the right ventricle, along with the increased preload due to the shunted blood. This will either cause right ventricular failure, or raise the pressure in the right side of the heart to equal or more than that in the left. Elevation of right atrial pressure to that of left atrial pressure would thus lead to diminishing or complete cessation of the shunt. If left uncorrected, there will be reversal of the shunt, known as Eisenmenger syndrome, resulting in clinical signs of cyanosis as the oxygen-poor blood form right side of the heart will mix with the blood in left side and reach the peripheral vascular system.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 4
Correct
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Which of the following is responsible for the activation of pepsinogen released in the stomach?
Your Answer: Acid pH and pepsin
Explanation:Pepsinogen is the inactive precursor of pepsin. Once secreted, it comes in contact with hydrochloric acid and pepsin, previously formed, and undergoes cleavage to form active pepsin.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 5
Correct
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Chest X-ray of a 45-year old gentleman with a week history of pleurisy showed a small pneumothorax with moderate-sized pleural effusion. Arterial blood gas analysis showed p(CO2) = 23 mmHg, p(O2) = 234.5 mmHg, standard bicarbonate = 16 mmol/l. What are we most likely dealing with?
Your Answer: Compensated respiratory alkalosis
Explanation:Normal pH with low p(CO2) and low standard bicarbonate could indicate either compensated respiratory alkalosis or a compensated metabolic acidosis. However, the history of hyperventilation for 5 days (pleurisy) favours compensated respiratory alkalosis. Compensated metabolic acidosis would have been likely in a diabetic patient with fever, vomiting and high glucose (diabetic ketoacidosis).
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 6
Correct
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Which type of contractions are responsible for the propulsion of chyme along the small intestine?
Your Answer: Segmentation
Explanation:Two major types of intestinal contractions are segmentation and peristalsis:
Segmentation occurs most frequently and primarily involves circular muscle. It is essentially a contraction of 2- or 3-cm long intestinal segments while the muscle on either side of it relaxes. Chyme in the segment is displaced in both directions. As the contracted segment relaxes, the previously relaxed segments on either side may contract. This efficiently mixes the chyme with the digestive secretions and exposes the mucosal absorptive surface to the luminal contents. It also serves a propulsive function and contributes to the movement of chyme.
Peristalsis is a propulsive wave of contraction that is initiated by intestinal distension. It is short lived and travels only a few centimetres before dying out. The combined effects of intestinal peristalsis and segmentation provide for both adequate mixing of the intestinal contents and slow, steady movement of chyme.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 7
Correct
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A 40 year old man suffered severe trauma following a MVA. His BP is 72/30 mmhg, heart rate of 142 beats/mins and very feeble pulse. He was transfused 3 units of blood and his BP returned to 100/70 and his heart rate slowed to 90 beats/min. What decreased after transfusion?
Your Answer: Total peripheral resistance
Explanation:The patient is in hypovolemic shock, he is transfused with blood, this fluid resuscitation will result in a decreased sympathetic discharge and adequate ventricular filling which will result in the decreases TPR with an increased CO and cardiac filling pressures
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 8
Incorrect
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The anatomical dead space in a patient with low oxygen saturation, is 125 ml, with a tidal volume of 500 ml and pa(CO2) of 40 mm Hg. The dead space was determined by Fowler's method. If we assume that the patient's lungs are healthy, what will his mixed expired CO2 tension [pE(CO2)] be?
Your Answer: 50 mmHg
Correct Answer: 30 mmHg
Explanation:According to Bohr’s equation, VD/VT = (pA(CO2) − pE(CO2))/pA(CO2), where pE(CO2) is mixed expired CO2 and pA(CO2) is alveolar CO2pressure. Normally, the pa(CO2) is virtually identical to pA(CO2). Thus, VD/VT = (pa(CO2)) − pE(CO2)/pa(CO2). By Fowler’s method, VD/VT= 0.25. In the given problem, (pa(CO2) − pE(CO2)/pa(CO2) = (40 − pE(CO2)/40 = 0.25. Thus, pE(CO2) = 30 mmHg. If there is a great perfusion/ventilation inequality, pE(CO2) could be significantly lower than 30 mm Hg, and the patient’s physiological dead space would exceed the anatomical dead space.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 9
Incorrect
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Extracellular body fluid as compared with intracellular body fluid:
Your Answer: Contains more amino acids
Correct Answer: Is relatively rich in glucose
Explanation:The percentages of body water contained in various fluid compartments add up to total body water (TBW). This water makes up a significant fraction of the human body, both by weight and by volume. Ensuring the right amount of body water is part of fluid balance, an aspect of homeostasis. The extracellular fluid (ECF) includes all fluids outside the cells. This fluid can be divided into three fluid departments: interstitial (in the tissue spaces) fluid, blood plasma and lymph, and specialised compartments called transcellular fluid. The extracellular fluid surrounds all the cells in the body and is in equilibrium with the intracellular fluid. So, its composition must remain fairly constant even though substances are passing into and out of the cells. The interstitial fluid, though called a fluid, is in a reality a gel-like composition made up of: water, proteoglycan molecules and collagen. The extracellular fluid constitutes 40% of total body water, with intracellular fluid making up the remaining 60%. It is relatively rich in glucose.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 10
Incorrect
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Which is a feature of the action of insulin?
Your Answer: Promotes phosphate release from bone
Correct Answer: Promotes protein synthesis
Explanation:Insulin is produced by the beta-cells of the islets of Langerhans in the pancreas. Its actions include:
– promoting uptake of glucose into cells
– glycogen synthesis (glycogenesis)
– protein synthesis
– stimulation of lipogenesis (fat formation).
– driving potassium into cells – used to treat hyperkaelamia.
Parathyroid hormone and activated vitamin D are the principal hormones involved in calcium/phosphate metabolism, rather than insulin.
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 11
Incorrect
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Which of the following organelles have the capacity to regenerate and spontaneously replicate?
Your Answer: Rough endoplasmic reticulum
Correct Answer: Mitochondrion
Explanation:A mitochondria is a membrane bound organelle found in eukaryotic cells. They are called the powerhouse of the cell and are the place where ATP is formed from energy generated through metabolism. They are capable of replication as well as repair and regeneration.
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This question is part of the following fields:
- General
- Physiology
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Question 12
Incorrect
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A 78-year-old diabetic man undergoes renal function tests. Which of the following substances will be the most accurate for measuring glomerular filtration rate (GFR)?
Your Answer: 51Cr-EDTA
Correct Answer: Inulin
Explanation:Glomerular filtration rate (GFR) is the volume of fluid filtered from the renal glomerular capillaries into the Bowman’s capsule per unit time. Clinically, this is often measured to determine renal function. Inulin was originally used as it is not reabsorbed by the kidney after glomerular filtration, therefore its rate of excretion is directly proportional to the rate of filtration of water and solutes across the glomerular filter. However, in clinical practice, creatinine clearance is used to measure GFR. Creatinine is an endogenous molecule, synthesised in the body, that is freely filtered by the glomerulus (but also secreted by the renal tubules in very small amounts). Creatinine clearance exceeds GFR due to creatinine secretion, and is therefore a close approximation of the GFR.
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This question is part of the following fields:
- Physiology
- Renal
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Question 13
Incorrect
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Intravenous diazepam was administered to a man who was brought to the emergency department with status epilepticus. He was administered 15 l/min oxygen via a reservoir bag mask. Blood investigations showed sodium = 140 mmol/l, potassium = 4 mmol/l and chloride = 98 mmol/l. His arterial blood gas analysis revealed pH 7.08, p(CO2)= 61.5 mmHg, p(O2) = 111 mmHg and standard bicarbonate = 17 mmol/l. This patient had:
Your Answer: Respiratory acidosis
Correct Answer: Mixed acidosis
Explanation:Acidosis with high p(CO2) and low standard bicarbonate indicates mixed acidosis. Lower p(O2) is due to breathing of 70% oxygen. The prolonged seizures lead to lactic acidosis and the intravenous diazepam is responsible for the respiratory acidosis. Treatment includes airway manoeuvres and oxygen, assisted ventilation if needed, and treatment with fluids.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 14
Incorrect
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Renin is secreted by pericytes in the vicinity of the afferent arterioles of the kidney from the juxtaglomerular cells. Plasma renin levels are decreased in patients with:
Your Answer: Salt restriction
Correct Answer: Primary aldosteronism
Explanation:Primary aldosteronism, also known as primary hyperaldosteronism or Conn’s syndrome, is excess production of the hormone aldosterone by the adrenal glands resulting in low renin levels. Most patients with primary aldosteronism (Conn’s syndrome) have an adrenal adenoma. The increased plasma aldosterone concentration leads to increased renal Na+ reabsorption, which results in plasma volume expansion. The increase in plasma volume suppresses renin release from the juxtaglomerular apparatus and these patients usually have low plasma renin levels. Salt restriction and upright posture decrease renal perfusion pressure and therefore increases renin release from the juxtaglomerular apparatus. Secondary aldosteronism is due to elevated renin levels and may be caused by heart failure or renal artery stenosis.
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This question is part of the following fields:
- Physiology
- Renal
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Question 15
Incorrect
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A 30 year old female presented in the emergency with an irregular pulse. Her ECG showed absent P-waves with irregular RR interval. What is the most likely diagnosis?
Your Answer: Ventricular fibrillation
Correct Answer: Atrial fibrillation
Explanation:Atrial fibrillation is one of the most common cardiac arrhythmias. It is often asymptomatic but may present with symptoms of palpitations, fainting, chest pain and heart failure. Characteristic findings are: absence of P-waves, unorganised electrical activity in their place, irregularity of RR interval due to irregular conduction of impulses to the ventricles and if paroxysmal AF is suspected, episodes may be documented with the use of Holter monitoring
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 16
Incorrect
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Which of the following is a likely consequence of severe diarrhoea?
Your Answer: A metabolic alkalosis
Correct Answer: A decrease in the sodium content of the body
Explanation:Diarrhoea can occur due to any of the numerous aetiologies, which include infectious, drug-induced, food related, surgical, inflammatory, transit-related or malabsorption. Four mechanisms have been implicated in diarrhoea: increased osmotic load, increased secretion, inflammation and decreased absorption time. Diarrhoea can result in fluid loss with consequent dehydration, electrolyte loss (Na+, K+, Mg2+, Cl–) and even vascular collapse. Loss of bicarbonate ions can lead to a metabolic acidosis.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 17
Incorrect
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Under normal conditions, what is the major source of energy of cardiac muscles?
Your Answer: Glucose
Correct Answer: Fatty acids
Explanation:Under basal conditions, most of the energy needed by cardiac muscle for metabolism is derived from fats (60%), 35% by carbohydrates, and 5% by ketones and amino acids. However, after intake of large amounts of glucose, lactate and pyruvate are mainly used. During prolonged starvation, fat acts as the primary source. 50% of the used lipids are sourced from circulating fatty acids.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 18
Incorrect
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The Henderson–Hasselbalch equation describes the derivation of pH as a measure of acidity. According to this equation, the buffering capacity of the system is at maximum when the number of free anions compared with undissociated acid is:
Your Answer: Exactly half
Correct Answer: Equal
Explanation:In 1908, Lawrence Joseph Henderson wrote an equation describing the use of carbonic acid as a buffer solution. Later, Karl Albert Hasselbalch re-expressed that formula in logarithmic terms, resulting in the Henderson–Hasselbalch equation. The equation is also useful for estimating the pH of a buffer solution and finding the equilibrium pH in acid–base reactions. Two equivalent forms of the equation are: pH = pKa + log10 [A–]/[HA] or pH = pKa + log10 [base]/[acid]. Here, pKa is − log10(Ka) where Ka is the acid dissociation constant, that is: pKa = –log10(Ka) = –log10 ([H3 O+][A–]/[HA]) for the reaction: HA + H2 O ≈ A– + H3 O+ In these equations, A– denotes the ionic form of the relevant acid. Bracketed quantities such as [base] and [acid] denote the molar concentration of the quantity enclosed. Maximum buffering capacity is found when pH = pKa or when the number of free anions to undissociated acid is equal and buffer range is considered to be at a pH = pKa ± 1.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 19
Incorrect
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The transmembrane proteins responsible for resting membrane potential of vascular smooth muscle cells was blocked by a drug. Which of the following transmembrane proteins were blocked by this drug?
Your Answer: Na+/K+ pump
Correct Answer: K+ channels
Explanation:The resting membrane potential is due to selective permeability of the membrane to potassium ions. The Na/K pump is responsible for the generation of a gradient across the membrane and it is due to the inherent ability of the K channels to allow diffusion back into the nerve at rest which charges the cells. In reality, the resting membrane potential is more positive because of small contributions by Na+ channels, Cl− channels and non-selective cation channels.
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This question is part of the following fields:
- General
- Physiology
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Question 20
Incorrect
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After surgery, a patient developed a stitch granuloma . Which leukocyte in the peripheral blood will become an activated macrophage in this granuloma?
Your Answer: Basophil
Correct Answer: Monocyte
Explanation:Monocytes are leukocytes that protect the body against infections and move to the site of infection within 8-12 hours to deal with it. They are produced in the bone marrow and shortly after being produced are released into the blood stream where they circulate until an infection is detected. When called upon they leave the circulation and transform into macrophages within the tissue fluid and thus gain the capability to phagocytose the offending substance. Monocyte count is part of a complete blood picture. Monocytosis is the state of excess monocytes in the peripheral blood and may be indicative of various disease states. Examples of processes that can increase a monocyte count include: • chronic inflammation • stress response • hyperadrenocorticism • immune-mediated disease • pyogranulomatous disease • necrosis • red cell regeneration.
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This question is part of the following fields:
- General
- Physiology
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Question 21
Incorrect
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C5a (a complement component) is a potent?
Your Answer: Opsonin
Correct Answer: Anaphylotoxin
Explanation:C5a is a strong chemoattractant as well as an anaphylotoxin and is involved in the recruitment of inflammatory cells such as neutrophils, eosinophils, monocytes, and T lymphocytes. It is also involved in activation of phagocytic cells, release of granule-based enzymes and generation of oxidants. All of which contribute to innate immune functions.
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This question is part of the following fields:
- General
- Physiology
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Question 22
Incorrect
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What is the normal duration of the ST segment?
Your Answer: 0.4 s
Correct Answer: 0.08 s
Explanation:The ST segment lies between the QRS complex and the T-wave. The normal duration of the ST segment is 0.08 s. ST-segment elevation or depression may indicate myocardial ischaemia or infarction.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 23
Incorrect
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A 25-year-old woman complains of generalised swelling and particularly puffiness around the eyes which is worst in the morning. Laboratory studies showed:
Blood urea nitrogen (BUN) = 30 mg/dl
Creatinine = 2. 8 mg/dl
Albumin = 2. 0 mg/dl
Alanine transaminase (ALT) = 25 U/l
Bilirubin = 1 mg/dl
Urine analysis shows 3+ albumin and no cells.
Which of the following is the most likely diagnosis?Your Answer: Interstitial renal disease
Correct Answer: Nephrotic syndrome
Explanation:Nephrotic syndrome is a disorder in which the glomeruli have been damaged, characterized by:
– Proteinuria (>3.5 g per 1.73 m2 body surface area per day, or > 40 mg per square meter body surface area per hour in children)
– Hypoalbuminemia (< 2,5 g/dl) – Hyperlipidaemia, and oedema (generalized anasarca).
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This question is part of the following fields:
- Physiology
- Renal
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Question 24
Correct
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what is the cause of a prolonged PT(prothrombin time)?
Your Answer: Liver disease
Explanation:PT measure the intrinsic pathway of coagulation. It determines the measure of the warfarin dose regime, liver disease and vit K deficiency status along with the clotting tendency of blood. PT measured factors are II,V,VII,X and fibrinogen. It is used along with aPTT which measure the intrinsic pathway.
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This question is part of the following fields:
- General
- Physiology
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Question 25
Incorrect
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In which of the following conditions will the oxygen-haemoglobin dissociation curve shift to the right?
Your Answer: Decrease in 2,3-diphosphoglycerate (DPG)
Correct Answer: Exercise
Explanation:The oxygen-haemoglobin dissociation curve plots saturated haemoglobin against the oxygen tension and is usually a sigmoid plot. Each molecule of haemoglobin can bind to four molecules of oxygen reversibly. Factors that can influence the binding include: pH, concentration of 2,3-diphosphoglycerate (2,3-DPG), temperature, type of haemoglobin molecules, and presence of toxins, especially carbon monoxide. Shape of the curve is due to interaction of bound oxygen molecules with the incoming molecules. The binding of first molecule is difficult, with easier binding of the second and third molecule and increase in difficulty with the fourth molecule – partly as a result of crowding and partly as a natural tendency of oxygen to dissociate.
Left shift of curve indicates haemoglobin’s increased affinity for oxygen (seen at lungs). Right shift indicates decreased affinity and is seen with an increase in body temperature, hydrogen ions, 2,3-diphosphoglycerate (DPG), carbon dioxide concentration and exercise. Under normal resting conditions in a healthy individual, the normal position of the curve is at a pH of 7.4. A shift in the position of the curve with a change in pH is called the Bohr effect. Left shift occurs in acute alkalosis, decrease in p(CO2), decrease in temperature and decrease in 2,3-DPG. The fetal haemoglobin curve is to the left of the adult haemoglobin to allow for oxygen diffusion across the placenta. The curve for myoglobin is even further to the left. Carbon monoxide has a much higher affinity for haemoglobin than oxygen does. Thus, carbon monoxide poisoning leads to hypoxia.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 26
Correct
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What is the normal amount of oxygen that is carried in the blood?
Your Answer: 20 ml oxygen/100 ml blood
Explanation:Normally, 100 ml of blood contains 15g haemoglobin and a single gram of haemoglobin can bind to 1.34 ml oxygen when 100% saturated. Thus, 15 × 1.34 = 20 ml O2/100 ml blood. The haemoglobin in venous blood that is leaving the tissues is about 75% saturated with oxygen, and hence it carries about 15 ml O2/100 ml venous blood. This implies that for each 10 ml of blood, 5 ml oxygen is transported to the tissues. With a p(O2) > 100 mm Hg, only 3 ml of oxygen is dissolved in every one litre of plasma. By increasing the pA(O2) by breathing 100% oxygen, one can add an extra amount of oxygen in the plasma, but the amount of oxygen carried by haemoglobin will not increase significantly as it is already > 95% saturated.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 27
Incorrect
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What is the normal duration of PR interval on an electrocardiogram of a healthy individual?
Your Answer: 0.001–0.002 s
Correct Answer: 0.12–0.20 s
Explanation:PR interval extends from the beginning of the P-wave until the beginning of the QRS complex. The normal duration of the PR interval is 0.12-0.20 s. It can be prolonged in first degree heart block, and reduced in Wolff-Parkinson-White syndrome.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 28
Incorrect
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Whilst snorkelling, a 30-year old gentleman has the respiratory rate of 10/min, tidal volume of 550 ml and an effective anatomical dead space of 250 ml. Which of the following will bring about a maximum increase in his alveolar ventilation?
Your Answer: A 2x increase in tidal volume
Correct Answer: A 2x increase in tidal volume and a shorter snorkel
Explanation:Alveolar ventilation = respiratory rate × (tidal volume − anatomical dead space volume). Increase in respiratory rate simply causes movement of air in the anatomical dead space, with no contribution to the alveolar ventilation. By use of a shorter snorkel, the effective anatomical dead space will decrease and will cause a maximum rise in alveolar ventilation along with doubling of tidal volume.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 29
Correct
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Which of the following organs is most likely to have dendritic cells?
Your Answer: Skin
Explanation:Dendritic cells are part of the immune system and they function mainly as antigen presenting cells. They are present in small quantities in tissues which are in contact in the external environment. Mainly in the skin and to a lesser extent in the lining of the nose, lungs, stomach and intestines. In the skin they are known as Langerhans cells.
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This question is part of the following fields:
- General
- Physiology
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Question 30
Incorrect
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A 40-year old lady with a flail chest due to trauma was breathing with the help of a mechanical ventilator in the ICU, and was heavily sedated on muscle relaxants. Due to sudden power failure, a nurse began to hand-ventilate the patient with a Ambu bag. What change will occur in the following parameters: (Arterial p(CO2), pH) in the intervening period between power failure and hand ventilation?
Your Answer: Increase, Increase
Correct Answer: Increase, Decrease
Explanation:Respiratory acidosis occurs due to alveolar hypoventilation which leads to increased arterial carbon dioxide concentration (p(CO2)). This in turn decreases the HCO3 –/p(CO2) and decreases pH. Respiratory acidosis can be acute or chronic. In acute respiratory acidosis, the p(CO2) is raised above the upper limit of normal (over 45 mm Hg) with low pH. However, in chronic cases, the raised p(CO2) is accompanied with a normal or near-normal pH due to renal compensation and an increased serum bicarbonate (HCO3 – > 30 mmHg). The given problem represents acute respiratory acidosis and thus, will show a increase in arterial p(CO2) and decrease in pH.
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This question is part of the following fields:
- Physiology
- Respiratory
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SESSION STATS - PERFORMANCE PER SPECIALTY
